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If $x_1$ and $x_2$ are the real roots of the equation $x^2-k x+c=0$, then the distance between the points $A\left(x_1, 0\right)$ and $B\left(x_2, 0\right)$ is
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Verified Answer
The correct answer is:
$\sqrt{k^2-4 c}$
Given, $x_1$ and $x_2$ are the roots of the equation
$$
\begin{array}{rlrl}
& \therefore & x^2-p x+c & =0 \\
& \text { and } & x_1+x_2 & =k \\
& x_1 x_2 & =-c
\end{array}
$$
$\therefore$ The distance between $A\left(x_1, 0\right)$ and $B\left(x_2, 0\right)$ is
$$
\begin{aligned}
A B & =\sqrt{\left(x_2-x_1\right)^2+0^2}=\left|x_2-x_1\right| \\
& =\sqrt{\left(x_1+x_2\right)^2-4 x_1 x_2} \\
& =\sqrt{k^2-4 \times 1 \times c} \\
& =\sqrt{k^2-4 c}
\end{aligned}
$$
$$
\begin{array}{rlrl}
& \therefore & x^2-p x+c & =0 \\
& \text { and } & x_1+x_2 & =k \\
& x_1 x_2 & =-c
\end{array}
$$
$\therefore$ The distance between $A\left(x_1, 0\right)$ and $B\left(x_2, 0\right)$ is
$$
\begin{aligned}
A B & =\sqrt{\left(x_2-x_1\right)^2+0^2}=\left|x_2-x_1\right| \\
& =\sqrt{\left(x_1+x_2\right)^2-4 x_1 x_2} \\
& =\sqrt{k^2-4 \times 1 \times c} \\
& =\sqrt{k^2-4 c}
\end{aligned}
$$
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